Hoarding Vault Tokens until the even re-runs

IIAlonditeII

This may not belong here, but I didn't know where it DOES go.

If I hoard tokens for a vault - say, The Hunt - until the next time that event runs, will the hoarded tokens still sit in the "expired" tokens bin and pull from an imaginary vault that resets after every pull, or will the be applied to the stash of tokens for the new vault?

TPF Alexis

It depends. If it's a regular event vault next time, they will pull from that vault. If the next run of that event is a new release, so it has a store instead of a vault, or and "...and friends" vault, you won't be able to use your hoarded tokens for that. I had a bunch of EotS tokens hanging around for over a year because of that at one point.

At least, that's how it's worked historically. There have been a couple of posts in the bugs forum recently about hoarded event tokens not being usable or flat out disappearing, so that might be changing.

IIAlonditeII

So in general, would the advice be "don't hoard vaults besides Tacos"?

TPF Alexis

I rarely hoard at all, but right now, yeah, I'm not sure hoarding vault tokens is a good idea. I'm also not sure it's a bad idea. There's a lot of uncertainty.

Neuromancer

There's no telling if an event will be retired. That said, this last time Enemy of the State ran, I had managed to save up enough tokens that I could clear the entire vault, and had 2 left over after the event ended. So, I was happy I hoarded. It just took a year between event runs to clear my hoard. :-P

Edit: I hoard not for the LT, though it is nice to pull, but to help complete diluted 4*s, and to complete necessary characters for announced SHIELD Training events.

helix72

There's only one reason to hoard vault tokens: to save in case a specific character is in it. Even then, given 4* dilution, the chances of it being the character you need and the color for that character you want is slim. So let's say all you care about is the top prize--the LT. In that case, it's better to draw as you go. Let me give an example:

Let's say I earn 10 tokens for a vault each time the event runs.

Option 1: I hoard, wait until the 8th run, then I have a 100% chance to get the one LT.

Option 2: Each event run I draw all those tokens and only hoard what is leftover if I get the LT in fewer than 10 draws. I ran 250 simulations of this (I really should run more to make it more credible, but it gives the picture), and I averaged 1.12 LTs drawing this way. Now, it is true there is a chance you go 0 for 80. But there is also a chance you go 8 for 80.

In the 250 simulations, here were the outcomes:

I got 0 LTs 81 times (32.4%)
I got 1 LT 95 times (38.0%)
I got 2 LTs 49 times (19.6%)
I got 3 LTs 18 times (7.2%)
I got 4 LTs 4 times (1.6%)
I got 5 LTs 2 times, 6 LTs once, and 7 and 8 zero times. I'm sure if I estimated this a million times it would be more than 0, but still a small chance

MadScientist

Never do simulations for something when you can calculate the probability directly.

This problem is  a model of "drawing from an urn without replacement". The number of items N=80 and the number of successes K=1 (the legendary token). The applicable probability distribution is the hypergeometric distribution. The mean number of successes after n tries is given by n*K/N, which is n/80 in our case.

Strategy 1: Draw 10 tokens on 8 different vaults. The expected number of legendary tokens is 1/8 * 8 = 1.

Strategy 2: Draw 80 tokens from one vault. The expected number of legendary tokens is 80/80 = 1.

As you can see, the expected number of LTs is exactly the same for the two strategies. 

TPF Alexis

Expected Value isn't actually a particularly good measure for this sort of thing, tho. Strategy 1 has massive variance, while Strategy 2 has none at all. It is still possible to calculate that spread exactly, but it's way more complicated than the Expected Value calculation, to the point where running simulations is probably significantly less work.

IIAlonditeII

MadScientist said:

Never do simulations for something when you can calculate the probability directly.

This problem is  a model of "drawing from an urn without replacement". The number of items N=80 and the number of successes K=1 (the legendary token). The applicable probability distribution is the hypergeometric distribution. The mean number of successes after n tries is given by n*K/N, which is n/80 in our case.

Strategy 1: Draw 10 tokens on 8 different vaults. The expected number of legendary tokens is 1/8 * 8 = 1.

Strategy 2: Draw 80 tokens from one vault. The expected number of legendary tokens is 80/80 = 1.

As you can see, the expected number of LTs is exactly the same for the two strategies. 

Except, the flaw with Strategy 1 is that you can draw... 0 in all 8 vaults. Strategy 2 cannot fail, and has the added benefit of leaving you with a well-stocked horde for the next vault in any event where you draw the token earlier than expected. IE, if you expected to draw it by 10 or so, you'd already have 70 tokens saved up for the next time.

helix72

IIAlonditeII said:

MadScientist said:

Never do simulations for something when you can calculate the probability directly.

This problem is  a model of "drawing from an urn without replacement". The number of items N=80 and the number of successes K=1 (the legendary token). The applicable probability distribution is the hypergeometric distribution. The mean number of successes after n tries is given by n*K/N, which is n/80 in our case.

Strategy 1: Draw 10 tokens on 8 different vaults. The expected number of legendary tokens is 1/8 * 8 = 1.

Strategy 2: Draw 80 tokens from one vault. The expected number of legendary tokens is 80/80 = 1.

As you can see, the expected number of LTs is exactly the same for the two strategies. 

Except, the flaw with Strategy 1 is that you can draw... 0 in all 8 vaults. Strategy 2 cannot fail, and has the added benefit of leaving you with a well-stocked horde for the next vault in any event where you draw the token earlier than expected. IE, if you expected to draw it by 10 or so, you'd already have 70 tokens saved up for the next time.

I will generally agree with @MadScientist regarding it being better to calculate the probability directly, however, it is not a strict hypergeometric distribution because as stated, we do not draw 10 tokens on 8 different vaults. We only draw 10 tokens if the first 9 do not draw the LT. If any of the first 9 draws are a success, then we immediately stop and save the extra tokens for the next vault. It is this wrinkle that makes direct calculation more complex and thus I went with a quick simulation as a reasonable approximation.

I agree with @IIAlonditeII that the volatility is greater, and while you can draw more tokens, you can also draw zero. They (not sure if you're a he or a she) also point out I did not consider that it would not take the full 80 tokens so you would have tokens left over. Unless someone feels like doing the conditional probabilities, I'll set up simulations over a longer range of vaults using 2 different strategies:

Strategy 1: pull up to however many tokens you have each vault, stopping once you get the LT, and applying any overage to the next vault

Strategy 2: wait until you have at least 80 tokens, pull until you get the LT, then stop and wait to pull again until you have at least 80 tokens. Lather, rinse, repeat.